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Hatshy [7]
3 years ago
10

Can someone solve under section two, numbers one and two?

Physics
2 answers:
Novay_Z [31]3 years ago
8 0

1. Height of cube =9cm

mass of cube= 3645 g

volume= length ×breadth× height

as cubic volume so length breadth and heights are same so

volume =height ×height ×height

V =9cm×9cm×9cm= 729cm³

density =mass/ volume

density =3645g/729cm³

density= 5g/cm³

2.  AS DENSITY = mass/volume

volume =mass/ desity

volume= 13.558g/(0.089g/l)

volume= 0.152L³

ser-zykov [4K]3 years ago
3 0

Density = mass/volume

#1). The mass is given. And knowing the length of a cube's side, you can calculate its volume.

Volume = mass/density

#2). The mass and density are both given.

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A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge
Troyanec [42]

Answer:

I_{l} =44.84 \mu A

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire =  1.30×10⁻² Ω  

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))

\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{V}{R}

I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}

I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))

I_{l} =44.84 \times 10^{-6}\A

I_{l} =44.84 \mu A

3 0
3 years ago
A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured
Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

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We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

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3 years ago
Redondear a dos decimales 12,4552​
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Answer:

12,46

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